## Reading for Class 7: Interference and Diffraction of Light Waves |

In case (a), the two waves match up exactly. We say they are "in-phase", since everywhere Wave 1 has a crest Wave 2 also has a crest, and everywhere Wave 1 has a trough, Wave 2 also has a trough, etc. When these two waves add, the result is a sine wave with twice the amplitude as one of the original waves. The waves in this case exhibit complete constructive interference.

In case (b), the waves are "180º out of phase", which means that they are off by one-half cycle. Every crest in Wave 1 matches a trough in Wave 2, and vice-versa. Adding these two waves yields an amplitude of zero. This case illustrates complete destructive interference.

Case (c) illustrates a situation somewhere between complete constructive and complete destructive interference. The two waves do not line up completely, but they do not cancel either. The result is a sine wave with an amplitude lower than twice the amplitude of one wave. The exact value of the amplitude will depend on how out-of-phase the two waves are. If the waves are nearly aligned, the amplitude of the sum will be larger than if the two waves are nearly canceling.

The difference between the distances traveled by the two waves is called the path difference, and it is represented by d on the diagram. The waves in case (a) needn't have traveled the exact same distance to line up; if one wave has traveled one wavelength more than the other has, the two waves would still line up. They would similarly align if one has traveled two or three or any integer number of wavelengths further than the other has. If the path difference is equal to an integer number of wavelengths, the two waves will experience complete constructive interference, and their amplitudes will add.

Similarly, the path difference for the waves in case (b) needn't be just l/2; if one has traveled three half-wavelengths more than the other has, they would still cancel. They would also cancel if one has traveled any other half-integer number of wavelengths further than the other has. If the path difference is equal to a half- integer number of wavelengths, the two waves will experience complete destructive interference, their amplitudes will cancel, and the sum will always have zero amplitude. Thus our conditions on the path difference d for interference are as follows:

d = ml
(m = 0, ±1, ±2, . . .) for constructive interference.

d = (m + 1/2)l
(m= 0, ±1, ±2, . . .) for destructive interference.

If the waves are light waves, spots of constructive
interference will be bright, and spots of destructive interference will
be dark. A screen represented by the black line in the above diagram
would therefore display regions of high-intensity light dimming into regions
of zero intensity before brightening into high-intensity regions again.
This collection of bright and dark regions is called an "interference pattern"
for light.

The pattern produced by a single slit is similar
to, but not identical to, the pattern produced by a double slit.
To distinguish between the two situations, we call the double-slit pattern
an interference pattern, and the single-slit pattern a diffraction pattern.
This distinction is somewhat misleading, since both effects depend upon
both diffraction (the spreading of light) and interference (combining of
light), but it is commonly used so you should be aware of it.

d = (*d* sin q)/2.44.

Here *d* is the diameter of the aperture,
and the factor of 2.44 comes from mathematical functions describing circular
geometry. As mentioned above, when the light rays cancel to produce
the first minimum, this path difference must be l/2.
Setting these two expressions for path difference equal to each other,
we get

l/2 = (*d* sin
q)/2.44

l = 2(*d* sin
q)/2.44
= (*d* sin q)/1.22.

For small angles, sin q is approximately equal to tan q, so

sin q ~ tan q
= *y*/*D* = 1.22 l/*d*.

Thus the location *y* of the first minimum
depends on the aperture size *d*, the distance *D* between the
aperture and the screen, and the wavelength l.

sin q £
1.22 l/*d*.

The bit must be no smaller than a circle of radius *r*, where

*r*/*D* =* *tan q*~
*1.22
l/*d*.
*r* ~ 1.22 l*D*/*d*

Theoretically, one could decrease this minimum
bit size in three different ways: using a laser with a shorter wavelength
l,
decreasing the distance *D* between the lens and the disk, or using
a lens with a larger diameter *d*. Moving the lens closer to
the disk or using a larger lens might seem the simplest ways to reduce
the diffraction pattern, but *D* and *d* are in practice tightly
constrained. Laser light passing through the lens will be focused
to the smallest spot at the focal point of the lens, so the disk-to-lens
distance *D* should equal the focal length of the lens. Thus
*D*
is determined by the lens and cannot be arbitrarily varied. Manufacturing
constraints on lenses also limit the width *d* of the lens.
Photographers are familiar with the
*f*-number, or *f*-stop of
lenses. This represents the ratio of focal length *D* over diameter
*d*
of the lens. Lenses with large *f*-numbers such as
*f*/8
or *f/*16 are relatively easy to produce. As the
*f*-number
decreases, the difficulty increases, and lenses with *f*-numbers less
than 1 are very rare. Thus the ratio *D*/*d* will be greater
than 1(and probably more like 5) for the mass-produced lenses in CD drives.
This leaves us with the option of reducing the laser wavelength as the
most promising way to decrease the minimum bit size in optical storage.

One of the reason's DVDs can hold more data than
CD-ROMs is that they use a shorter wavelength of light, so the resulting
diffraction pattern on the disk is smaller. Bits may then each occupy
less space, so more bits fit on a disk. Your handout from Ron White's
book discusses additional reasons that DVDs store more data than CD-ROMs.

- |
Light passing through a circular aperture
is diffracted and produces a ringed diffraction pattern. |

- | For two objects to be resolved with such light,
they must not both lie within the central maximum. |

- |
The size y of the central maximum can
be reduced in three ways: decreasing the distance D between
the slit and the screen, increasing the diameter d of the aperture,
and decreasing the wavelength l. |

- |
In practical applications (like CD-ROMs),
the aperture size d and distance between aperture and screen (data)
D
are severly constrained. Thus adjusting the wavelength becomes the
most effective way to decrease the size of the diffraction pattern and
thereby resolve smaller objects. |

*How Computers Work*, by Ron White.

The HowStuff Works Web site's CD page

Notes by Professor Kelin J. Kuhn for his EE courses: Audio Compact Disk - an Introduction, and Audio Compact Disk - Reading and Writing the Data. (These are from a link at HowStuffWorks.) These get a bit technical at times, but they present a much more complete picture of CDs than White's book does.

Any introductory physics text, such as *Fundamentals of Physics*
by Halliday, Resnick and Walker.

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© 2000-2002 Doris Jeanne Wagner. All Rights Reserved.**